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Simora [160]
3 years ago
9

Many times 4 can go into a 100?

Mathematics
1 answer:
Alex787 [66]3 years ago
3 0
4 × ? = 100

? = 100 : 4

? = <u>25</u>
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Un carro recorre 784 kilometros en una hora expresar la rapidez en m/sg
9966 [12]

Answer:

217.7777 m/s

Step-by-step explanation:

1km = 1000 metros

784 km = 784*1000 = 784000 metros

1 hora = 3600 segundos

entonces:

784 kilómetros en una hora = 784km/h = 784000m/3600ses

= 217.7777 m /s

es decir que recorre en cada segundo:

217.7777 metros.

Tarea relacionada:

https://brainly.lat/tarea/16056884

4 0
4 years ago
A polynomial p(x) is divided by a binomial x-a the remainder is 0
schepotkina [342]
If this is the case, then you know by the polynomial remainder theorem that x-a is a factor of p(x), so there is some polynomial q(x) such that

p(x)=(x-a)q(x)
3 0
3 years ago
Suppose the graph of a cubic polynomial function
ivann1987 [24]

Answer:

(1/6)(x-2)(x-3)(x-5)

Step-by-step explanation:

7 0
3 years ago
If y=14 then what is 1.8+0.5(y+6)-2 to the third power
pychu [463]
3.8 because 14+6=20 and .5*20=10 so then you do PEMDAS which means parenthesis, exponents, multiply, divide, add, subtract left to right. So the equation now is 1.8+10-8 so you would do 1.8+10=11.8 and 11.8-8=3.8. In PEMDAS you do the parenthesis first then the exponents and then multiply and divide and then when there is only add and subtract left in the equation you do it left to right. 
7 0
3 years ago
Find an equation of the set of all points equidistant from the points A(−3, 6, 3) and B(4, 1, −1). Describe the set.
valentinak56 [21]

Answer:

See below

Step-by-step explanation:

I will describe this set in R³. Let P=(x,y,z) be a point equidistant to A and B, that is, the distance from P to A is equal to the distance from P to B.

First, using the usual distance formula, the distance from P to A is equal to d(P,A)=\sqrt{(x-(-3))^2+(y-6)^2+(z-3)^2}=\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}

On the other hand, the distance form P to B is equal to d(P,B)=\sqrt{(x-4)^2+(y-1)^2+(z-(-1))^2}=\sqrt{(x-4)^2+(y-1)^2+(z+1)^2}

P is equidistant from A and B if and only if P satisfies the equation d(P,A)=d(P,B), that is,

\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}=\sqrt{(x-4)^2+(y-1)^2+(z+-1)^2}

Take the square in both sides of this equation to get

(x+3)^2+(y-6)^2+(z-3)^2=(x-4)^2+(y-1)^2+(z+1)^2

(x+3)^2-(x-4)^2+(y-6)^2-(y-1)^2+(z-3)^2-(z+1)^2=0

You can simplify using difference of squares and multiplying like this:

(7)(2x-1)+(-5)(2y-7)+(-4)(2z-2)=0

14x-10y-8z+36=0

which is the equation of a plane.

4 0
4 years ago
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